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Lời giải của bạn Nhật Linh đúng rồi, tuy nhiên cần thêm điều kiện để A có nghĩa: \(x\ne\pm2\)
1/ a, \(A=\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
\(=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{3x-x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{1}{x}\)
Vậy \(A=x\)
b/ Khi \(x=\dfrac{1}{2}\Leftrightarrow A=\dfrac{1}{\dfrac{1}{2}}=2\)
Vậy...
2/a,
\(A=\dfrac{5x+2}{3x^2+2x}+\dfrac{-2}{3x+2}\)
\(=\dfrac{5x+2}{x\left(3x+2\right)}-\dfrac{2x}{x\left(3x+2\right)}\)
\(=\dfrac{5x+2-2x}{x\left(3x+2\right)}\)
\(=\dfrac{3x+2}{x\left(3x+2\right)}\)
\(=\dfrac{1}{x}\)
Vậy....
b/ Với \(x=\dfrac{1}{3}\Leftrightarrow A=\dfrac{1}{\dfrac{1}{3}}=3\)
Vậy..
1.
a) \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
b) \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Bài 1:
a, \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
Vậy \(x=-4\) hoặc \(x=-1\)
b, \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=-2\)
\(A=\left(\dfrac{x+y}{y}+\dfrac{2y}{x-y}\right)\cdot\dfrac{-\left(x-y\right)}{x^2+y^2}+\left(\dfrac{2x^2+2-2x^2+x}{2\left(2x-1\right)}\right)\cdot\dfrac{1-2x}{x+2}\)
\(=\dfrac{x^2-y^2+2y^2}{y\left(x-y\right)}\cdot\dfrac{-\left(x-y\right)}{x^2+y^2}+\dfrac{x+2}{2\left(2x-1\right)}\cdot\dfrac{-\left(2x-1\right)}{x+2}\)
\(=\dfrac{-1}{y}+\dfrac{-1}{2}=\dfrac{-2-y}{2y}\)
a: ĐKXĐ: x<>3; x<>-3; \(x\ne-5\pm\sqrt{34}\)
b: \(=\dfrac{x^2+5x+6+5x-15}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{2x\left(x-3\right)\left(x+3\right)}{x^2+10x-9}\)
=2x
c: Khi x=1/2 thì A=2*1/2=1
a) \(A=\left(\dfrac{x-2}{x^2-1}-\dfrac{x+2}{x^2+2x+1}\right).\dfrac{x^2-1}{2}\left(ĐKXĐ:x\ne\pm1\right)\)
\(A=\left[\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+2}{\left(x+1\right)^2}\right].\dfrac{\left(x+1\right)\left(x-1\right)}{2}\)
\(A=\left[\dfrac{\left(x-2\right)\left(x+1\right)-\left(x+2\right)\left(x-1\right)}{\left(x+1\right)^2\left(x-1\right)}\right].\dfrac{\left(x+1\right)\left(x-1\right)}{2}\)
\(A=\left[\dfrac{x^2+x-2x-2-\left(x^2-x+2x-2\right)}{\left(x+1\right)^2\left(x-1\right)}\right].\dfrac{\left(x+1\right)\left(x-1\right)}{2}\)
\(A=\left[\dfrac{x^2-x-2-x^2-x+2}{\left(x+1\right)^2\left(x-1\right)}\right].\dfrac{\left(x+1\right)\left(x-1\right)}{2}\)
\(A=\dfrac{-2x}{\left(x+1\right)^2\left(x-1\right)}.\dfrac{\left(x+1\right)\left(x-1\right)}{2}\)
\(A=\dfrac{-2x}{2\left(x+1\right)}\)
P/s: câu b bn tự làm nha
a.
\(A=\left(\dfrac{x-2}{x^2-1}-\dfrac{x+2}{x^2+2x+1}\right)\dfrac{x^2-1}{2}\)
\(A=\left(\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+2}{\left(x+1\right)^2}\right).\dfrac{x^2-1}{2}\)
\(A=\left(\dfrac{\left(x-2\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)^2}-\dfrac{\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)^2}\right).\dfrac{x^2-1}{2}\)
\(A=\dfrac{x^2-x-2-x^2-x+2}{\left(x-1\right)\left(x+1\right)^2}.\dfrac{x^2-1}{2}\)
\(A=\dfrac{-2x}{\left(x-1\right)\left(x+1\right)^2}.\dfrac{\left(x-1\right)\left(x+1\right)}{2}\)
\(A=\dfrac{-x}{x+1}\)